Theorem. If the sum of the interior angles of every triangle equals straight angle, then the parallel postulate is true, i.e., in the plane determined by a line and a point outwards it there is exactly one line through the point which does not intersect the line.

Proof. We consider a line a and a point B not belonging to a. Let B⁢A be the normal line of a
(with A∈a) and b be the normal line of B⁢A through the point B. By the supposition of the theorem, b does not intersect a.

We will show that in the plane determined by the line a and the point B, there are through B no other lines than b not intersecting the line a. For this purpose, we choose through B a line b′ which differs from b; let the line b′ form with B⁢A an acute angleβ.

We determine on the line a a point A1 such that A⁢A1=A⁢B. By the supposition of the theorem, in the isosceles right triangleB⁢A⁢A1 we have

α1=:∠AA1B=π4=π22.

Next we determine on a a second point A2 such that
A1⁢A2=A1⁢B. By the supposition of the theorem, in the isosceles triangleB⁢A1⁢A2 we have

α2=:∠AA2B=α12=π23.

We continue similarly by forming isosceles triangles using the points A3, A3, …, An of the line a such that

A2⁢A3=B⁢A2,A3⁢A4=B⁢A3,…,An-1⁢An=B⁢An-1.

Then the acute angles being formed beside the points are

α3=π24,α4=π25,…,αn=π2n+1.

They form a geometric sequence with the common ratior=12. When n is sufficiently great, the member αn is less than any given positive angle. As we have so much triangles B⁢An-1⁢An that
αn<π2-β, then

∠⁢A⁢B⁢An=π2-αn>β.

Then the line b′ falls after penetrating B into the inner territory of the triangle A⁢B⁢An. Thereafter it must leave from there and thus intersect the side A⁢An of this triangle. Accordingly, b′ intersects the line a.

The above reasoning is possible for each line b′≠b through B. Consequently, the parallel axiom is in force.